Solve for 0
please show how to do this i need to learn it
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The best thing to do is make all the trig functions the same function
Lets start by squaring both sides
(1 - sinx)² = (3cosx)²
1 - 2sinx + sin²x = 9cos²x
Unfortunately, by squaring we may have inadvertently introduced false solutions. We must check for this in the end.
Use Pythagorean Identity
1 - 2sinx + sin²x = 9(1 - sin²x)
1 - 2sinx + sin²x = 9 - 9sin²x
-8 - 2sinx + 10sin²x = 0
We are left with a quadratic equation in terms of sin(x).
10sin²x - 2sinx - 8 = 0
5sin²x - sinx - 4 = 0
Factor
(5sin x + 4)(sin x - 1) = 0
And we have solutions for each factor
sin(x) = 1
sin(x) = -4/5
For the first, sin(x) = 1, we have the solutions:
x = ½π + 2πn
Or just
x = ½π, since we are trying to stay in the narrow domain
Checking with the original expression, this solution does work.
For the second, sin(x) = -4/5, we have the solutions:
x ≈ -0.927295218... + 2πn , which ends up being a transcendental number.
Or x ≈ 5.35589... to stay in the domain.
It too does work with the original expression.
Lets start by squaring both sides
(1 - sinx)² = (3cosx)²
1 - 2sinx + sin²x = 9cos²x
Unfortunately, by squaring we may have inadvertently introduced false solutions. We must check for this in the end.
Use Pythagorean Identity
1 - 2sinx + sin²x = 9(1 - sin²x)
1 - 2sinx + sin²x = 9 - 9sin²x
-8 - 2sinx + 10sin²x = 0
We are left with a quadratic equation in terms of sin(x).
10sin²x - 2sinx - 8 = 0
5sin²x - sinx - 4 = 0
Factor
(5sin x + 4)(sin x - 1) = 0
And we have solutions for each factor
sin(x) = 1
sin(x) = -4/5
For the first, sin(x) = 1, we have the solutions:
x = ½π + 2πn
Or just
x = ½π, since we are trying to stay in the narrow domain
Checking with the original expression, this solution does work.
For the second, sin(x) = -4/5, we have the solutions:
x ≈ -0.927295218... + 2πn , which ends up being a transcendental number.
Or x ≈ 5.35589... to stay in the domain.
It too does work with the original expression.